Question
A sum becomes Rs.4500 after two years and Rs.6750 after four years at compound interest. The sum is
Answer: Option C
Answer: (c)P$(1 + r/100)^2$ = 4500 ...(i)P$(1 + r/100)^4$ = 6750 .....(ii)On dividing equation (ii) by equation (i), we get$(1 + r/100)^2 = 6750/4500$From equation (i),P × $6750/4500$ = 4500P = ${4500 × 4500}/6750$ = Rs.3,000Using Rule 7(ii),Here, b - a = 4 - 2 = 2B = Rs.6750, A = Rs.4500R% = $[(B/A)^{1/2} - 1]$ × 100%= $[(6750/4500)^{1/2} - 1] × 100%$= $[(3/2)^{1/2} - 1] ×$ 100%$(3/2)^{1/2} = 1 + R/100$$3/2 = (1 + R/100)^2$A = P$(1 + R/100)^2$4500 = P × $3/2$ ⇒ P = Rs.3000
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Answer: (c)P$(1 + r/100)^2$ = 4500 ...(i)P$(1 + r/100)^4$ = 6750 .....(ii)On dividing equation (ii) by equation (i), we get$(1 + r/100)^2 = 6750/4500$From equation (i),P × $6750/4500$ = 4500P = ${4500 × 4500}/6750$ = Rs.3,000Using Rule 7(ii),Here, b - a = 4 - 2 = 2B = Rs.6750, A = Rs.4500R% = $[(B/A)^{1/2} - 1]$ × 100%= $[(6750/4500)^{1/2} - 1] × 100%$= $[(3/2)^{1/2} - 1] ×$ 100%$(3/2)^{1/2} = 1 + R/100$$3/2 = (1 + R/100)^2$A = P$(1 + R/100)^2$4500 = P × $3/2$ ⇒ P = Rs.3000
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